Question

An alpha particle travels at a velocity of magnitude 820 m/s through a uniform magnetic field of magnitude 0.038 T. (An alpha particle has a charge of charge of +3.2 × 10^-19 C and a mass 6.6 × 10^-27 kg) The angle between the particle's direction of motion and the magnetic field is 69°.

What is the magnitude of

(a) the force acting on the particle due to the field,

(b) the acceleration of the particle due to this force?

(c) Does the speed of the particle increase, decrease, or remain the same?

Answer 1

2.

Force on a charged particle moving in a magnetic field is given by:

Fm = q*VxB = q*V*B*sin theta

Given that:

q = charge on alpha particle = 3.2*10^-19 C

V = Velocity of particle = 820 m/s

B = magnetic field = 0.038 T

theta = angle between direction of motion and magnetic field = 69 deg

So,

F = 3.2*10^-19*820*0.038*sin 69 deg

F = 9.31*10^-18 N = Force acting on the particle due to the field

Part B.

Using Force balance:

F_net = m*a

m = mass of alpha particle = 6.6*10^-27 kg

So,

a = F_net/m = 9.31*10^-18/(6.6*10^-27)

a = 1.41*10^9 m/s^2 = acceleration of the particle due to this force

Part C.

Since magnetic force is always in perpendicular direction of velocity of particle, So net work-done by magnetic force will be zero, which means speed of particle will remain constant.

Let me know if you've any query.